Does series $\sum _{n=1}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge? Does series 
$\sum _{n=2}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge?
My idea is to apply the Cauchy test, but I dont know how to simplify it next.
Thanks
 A: Plenty of good answers already given: all right, we may go for the overkill. By Frullani's theorem the logarithm function has a nice integral representation:
$$ \log\frac{m}{n} = \int_{0}^{+\infty}\frac{e^{-nx}-e^{-mx}}{x}\,dx $$
hence it follows that:
$$S=\sum_{n\geq 2}\log\frac{n(n+2)}{(n+1)^2} = \int_{0}^{+\infty}\sum_{n\geq 2}\frac{2e^{-(n+1)x}-e^{-nx}-e^{-(n+2)x}}{x}\,dx $$
simplifying, by geometric series, to:
$$ S = \int_{0}^{+\infty}\frac{e^{-3x}-e^{-2x}}{x}\,dx = \color{red}{\log\frac{2}{3}}.$$
A: $$\color{red}{-\ln\left(\frac{27}{16}\right)}=\int_1^\infty\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)dn \color{red}{< \sum _{n=2}^{\infty}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right) <} \sum _{n=2}^{\infty}\ln\left(\frac{\left(n+1\right)^2}{\left(n+1\right)^2}\right)= \color{red}{0}$$
Inequalities follow from the fact that our function is strictly increasing.
Note that this bounds the answer decently well as a bonus.
A: Observe that the partial sum
$$
\sum_{n=2}^k\ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\sum_{n=2}^k \left(\ln(n)-2\ln(n+1)+\ln(n+2)\right)
$$
We can then rearrange the sum to get that the partial sum is
$$
\sum_{n=2}^k\ln(n)-2\sum_{n=2}^k\ln(n+1)+\sum_{n=2}^k\ln(n+2)=\sum_{n=2}^k\ln(n)-2\sum_{n=3}^{k+1}\ln(n)+\sum_{n=4}^{k+2}\ln(n).
$$
This simplifies (by telescoping) to
$$
\ln(2)-\ln(3)-\ln(k+1)+\ln(k+2)=\ln\left(\frac{2}{3}\right)+\ln\left(\frac{k+2}{k+1}\right).
$$
As $k$ approaches infinity, $\frac{k+2}{k+1}$ approaches $1$ (and $\ln(1)=0$), so the limit exists and is $\ln\left(\frac{2}{3}\right)$.
A: Hint:
$$\frac{n(n+2)}{(n+1)^2}=1-\frac1{(n+1)^2}$$ so that 
$$\log\frac{n(n+2)}{(n+1)^2}$$ is asymptotically
$$-\frac1{(n+1)^2}$$ and the series converges (like $\zeta(2)$).
A: Hint: $ \ln (1+u)/u\to 1$ as $u \to 0,$ and $n(n+2)/(n+1)^2 = 1 -1/(n+1)^2.$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\sum_{n = 2}^{\infty}\ln\pars{n\bracks{n + 2} \over \bracks{n + 1}^{\,2}} =
\sum_{n = 3}^{\infty}\ln\pars{1 - {1 \over n^{2}}} =
\sum_{n = 3}^{\infty}\bracks{-2\int_{0}^{1}{t \over n^{2} - t^{2}}\,\dd t}
\\[5mm] = &\
-2\int_{0}^{1}t\sum_{n = 3}^{\infty}{1 \over n^{2} - t^{2}}\,\dd t =
-\int_{0}^{1}\sum_{n = 3}^{\infty}\pars{{1 \over n - t} - {1 \over n + t}}
\,\dd t
\\[5mm] = &\
-\int_{0}^{1}\sum_{n = 0}^{\infty}
\pars{{1 \over n + 3 - t} - {1 \over n + 3 + t}}\,\dd t
\\[5mm] = &\
-\int_{0}^{1}\bracks{\Psi\pars{3 + t} - \Psi\pars{3 - t}}\,\dd t\qquad\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] = &\
\left.\vphantom{\LARGE A}-\ln\pars{\vphantom{\Large A}\Gamma\pars{3 + t}\Gamma\pars{3 - t}}
\right\vert_{\ 0}^{\ 1}\qquad\qquad\qquad\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] = &\
-\ln\pars{\Gamma\pars{4}\Gamma\pars{2} \over \Gamma^{2}\pars{3}} =
-\ln\pars{6 \times 1 \over 2^{2}} = \bbx{\ds{\ln\pars{2 \over 3}}}
\end{align}
A: Using the identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ we have $$\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}=\prod_{n\geq0}\frac{\left(n+2\right)\left(n+4\right)}{\left(n+3\right)^{2}}=\frac{\Gamma^{2}\left(3\right)}{\Gamma\left(2\right)\Gamma\left(4\right)}=\frac{2}{3}.$$ so $$\sum_{n\geq2}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\color{red}{\log\left(\frac{2}{3}\right)}.$$
